WEBVTT
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GILBERT STRONG: Well, you
see spring has finally
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come to Boston.
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My sweater is gone and it's
April the 16th, I think.
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It's getting late spring.
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So today, this video is the
nice case, constant coefficient,
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linear equations, and the right
hand side is an exponential.
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Those are the best.
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And we've seen that before.
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In fact, let me extend, we saw
it for first order equations,
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here it is for second
order equations,
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and it could be an
nth order equation.
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We could have the nth derivative
and all lower derivatives.
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The first derivative,
the function itself,
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0-th derivative
with coefficients,
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constant coefficients,
equalling e
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to the st. That's
what makes it easy.
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And what do we do when the
right hand side is e to the st?
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We look for a
solution, a multiple
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of e to the st.
Capital Y times e
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to the st, that's going to work.
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We just plug into the equation
to find that transfer function
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capital Y. Can I just do that?
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I'll do it for the nth degree.
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Why not?
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So will I plug this in for y,
every derivative brings an s.
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Capital Y is still
there, the exponential
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is going to be still
there, and then
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there are all the s's that
come down from the derivatives,
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and s's from the nth derivative.
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One s from the first derivative,
no s from the constant term.
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Do you see that
equation is exactly
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like what we had before with
as squared plus Bs plus C.
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We have quadratic equation,
the most important case.
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Now, I'm including that with
any degree equation, nth degree
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equation.
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And what's the solution for Y?
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Because e to the st
cancels e to the st.
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That whole thing equals 1.
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I divide by this
and I get Y equal 1
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over that key polynomial.
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It's a nth degree polynomial.
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And the 1 over it is called
the transfer function.
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And that transfer function
transfers the input--
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e to the st-- to the
output-- Ye to the st.
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It gives the
exponential response.
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Very nice formula.
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Couldn't be better.
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And you remember for
second degree equations,
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our most important case is
as squared Bs plus C. That's
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the solution almost every time.
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But one thing can go wrong.
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One thing can go wrong.
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Suppose for the particular
s, the particular exponent,
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in the forcing function, suppose
that s in the forcing function
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is also one of the s's
in the no solutions.
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You remember the no solutions,
there are two s's-- s1 and s2--
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that make this 0,
that make that 0.
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Those are the s1 and s2 that
go into the no solutions.
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Now, if the forcing s is one
of those no solution s's, we
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have a problem.
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Because this is 1 over 0 and
we haven't got an answer yet.
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1 over 0 has no meaning.
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So I have to, this
is called resonance.
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Resonance is when
the forcing exponent
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is one of the no exponents
that make this 0.
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And there are two of those
for second order equations
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and there will be n different
s's-- s1, s2, up to sn--
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for nth degree equations.
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Those special s's, I
could also call them
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poles of the transfer function.
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The transfer function has
this in the denominator
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and when this is 0,
that identifies a pole.
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So the s1, s2, to
sn are the poles
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and we hope that this s is not
one of those, but it could be.
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And if it is, we
need a new formula.
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So that's the only
remaining case.
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This is a completely
nice picture.
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We just need this last
case with some resonance
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when s equals say s1.
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I'll just pick s1 and
when A, I know that As 1
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squared plus Bs 1 plus C is 0.
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So Y would be 1 over 0.
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And we can't live with that.
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I've written here for the
second degree equation
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same possibility
for the nth degree,
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An s1 to the nth
plus A0 equals 0.
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That would be a
problem of resonance.
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In the nth degree equation, this
gives us resonance, you see,
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because remember, the no
solutions were e to the s1t
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was a no solution.
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If I plug-in e to the s1t,
the left side will give 0.
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So I can't get for equal to a
forcing term on the right side.
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I need a new solution.
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I need a new y of t.
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Can I tell you what it is?
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It's a typical case of
L'Hopital's rule from calculus
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when we approach
this bad situation,
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and we are getting a 1 over 0.
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Well, you'll see a 0 over 0
and that's L'Hopital's aim for.
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So where do we start?
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A Y particular
solution is this e
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to the st over this As
squared plus Bs plus C. Right.
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That's our particular solution.
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If it works.
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Resonance is the case when this
doesn't work because that's 0.
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Now, that's a
particular solution.
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I'll subtract off a no solution.
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I could do that.
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I still have a solution.
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So I subtract off e to the s1t.
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So S1 is, e to the
s1t is a no solution.
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This is what I would call
a very particular solution.
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It's very particular because
a t equals 0, it's 0.
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Do you see that, you see
what's happening here.
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The question, resonance
happens when s approaches s1.
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Resonance is s
equal to, resonance
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itself is at the thing.
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Now, we let s sneak
up on s1 and we ask
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what happens to that formula.
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You see, we're sneaking
up on resonance.
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At resonance, when s
equals s1, that will be 0
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and that will be 0.
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That's our problem.
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So approach it and you
end up with the derivative
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of this divided by the
derivative of this.
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Do you remember L'Hopital?
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It was a crazy rule in calculus,
but here it's actually needed.
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So as s goes to s1
this goes to 0 over 0,
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so I have to take
derivative over derivative.
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So let me write the answer.
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Y resonant.
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Can I call this the resonant
solution when s equals s1.
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And what does it equal?
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Well, I take the s
derivative of this
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and divide it by the
s derivative of that.
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Derivative over derivative.
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The s derivative of
that is te to the st.
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And the s derivative
of this is 2As plus B.
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And now, I have derivative
over derivative,
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I can let s go to s1.
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So s goes to s1, this goes
to s1, and I get an answer.
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The right answer.
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This is the correct
solution and you notice
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everybody spots this t factor.
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That t factor is a
signal to everyone
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that we're in a
special case when
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two things happen to be equal.
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Here the two things
are the s and the s1.
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So that will work.
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So do you see the
general picture?
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It's always this te
to the st, t above.
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And down below we have the
derivative of this polynomial
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at s equal s1.
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You know it's theoretically
possible that we
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could have double resonance.
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We would have double
resonance if that thing is 0.
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If s1 was a double root, if
s1 was a double root, then,
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well, that's just absurd,
but it could happen.
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Then not only is that
denominator 0, but after one
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use of L'Hopital, so we
have to drag L'Hopital back
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from the hospital
and say do it again.
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So we would have a
second derivative.
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I won't write down that solution
because it's pretty rare.
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So what has this video done?
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Simply put on record
the simplest case
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possible with a forcing
function, e to the st.
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And above all, we've identified
this transfer function.
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And let me just anticipate that
if we need another way to solve
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these equations, instead
of in the t domain,
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we could go to the Laplace
transform in the s domain.
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We could solve it
in the s domain.
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And this is exactly
what we'll meet when
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we take the Laplace transform.
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That will be the
Laplace transform,
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which we have to deal with.
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So that transfer function
is a fundamental,
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this polynomial tells us, its
roots tell us the frequencies,
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s1, s2, and the no solutions.
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And then 1 over that tells
us the right multiplier
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in the force solution.
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So constant coefficients,
exponential forcing,
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the best case possible.
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Thank you.